A chemist has two solutions: one containing 20% alcohol and another containing 50% alcohol.
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A chemist has two solutions: one containing 20% alcohol and another containing 50% alcohol. How many liters of each solution must be mixed to obtain 30 liters of a solution that is 40% alcohol?
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Set up two equations: x + y = 30 (total volume) and 0.20x + 0.50y = 0.40(30) (alcohol content balance). Solving this system yields x = 10 liters of 20% solution and y = 20 liters of 50% solution. Verification: 0.20(10) + 0.50(20) = 2 + 10 = 12 liters of pure alcohol in 30 liters total = 40%.
Step-by-step Derivation:
Let x = liters of 20% solution and y = liters of 50% solution.
Equation 1 (volume): x + y = 30
Equation 2 (alcohol content): 0.20x + 0.50y = 0.40(30) → 0.20x + 0.50y = 12
From Equation 1: x = 30 - y
Substitute into Equation 2:
0.20(30 - y) + 0.50y = 12
6 - 0.20y + 0.50y = 12
6 + 0.30y = 12
0.30y = 6
y = 20 liters
Therefore: x = 30 - 20 = 10 liters
Verification:
- Total volume: 10 + 20 = 30 ✓
- Alcohol content: 0.20(10) + 0.50(20) = 2 + 10 = 12 liters
- Percentage: 12/30 = 0.40 = 40% ✓